What is the volume of 45.6 g of silver if the density of silver is 10.5 g/mL?

Sep 24, 2016

Simply solve the quotient: $\frac{45.6 \cdot g}{10.5 \cdot g \cdot m {L}^{-} 1}$ $\cong$ $4 \cdot m L$
$\rho , \text{density}$ $=$ $\text{Mass"/"Volume}$
And thus $\text{Volume}$ $=$ $\frac{\text{Mass}}{\rho}$ $=$ $\frac{45.6 \cdot \cancel{g}}{10.5 \cdot \cancel{g} \cdot m {L}^{-} 1}$ $=$ ??mL.
This is consistent dimensionally in that $\frac{1}{m {L}^{-} 1}$ $=$ $\frac{1}{\frac{1}{m L}}$ $=$ $1 \cdot m L$ as required.